251 lines
8.2 KiB
Python
251 lines
8.2 KiB
Python
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# -*- coding: utf-8 -*-
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# Copyright (c) 2018, Brandon Nielsen
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# All rights reserved.
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#
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# This software may be modified and distributed under the terms
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# of the BSD license. See the LICENSE file for details.
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import datetime
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from aniso8601.exceptions import DayOutOfBoundsError, ISOFormatError, \
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WeekOutOfBoundsError, YearOutOfBoundsError
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from aniso8601.resolution import DateResolution
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def get_date_resolution(isodatestr):
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#Valid string formats are:
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#
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#Y[YYY]
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#YYYY-MM-DD
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#YYYYMMDD
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#YYYY-MM
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#YYYY-Www
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#YYYYWww
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#YYYY-Www-D
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#YYYYWwwD
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#YYYY-DDD
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#YYYYDDD
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if isodatestr.startswith('+') or isodatestr.startswith('-'):
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raise NotImplementedError('ISO 8601 extended year representation not supported.')
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if isodatestr.find('W') != -1:
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#Handle ISO 8601 week date format
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hyphens_present = 1 if isodatestr.find('-') != -1 else 0
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week_date_len = 7 + hyphens_present
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weekday_date_len = 8 + 2 * hyphens_present
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if len(isodatestr) == week_date_len:
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#YYYY-Www
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#YYYYWww
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return DateResolution.Week
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elif len(isodatestr) == weekday_date_len:
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#YYYY-Www-D
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#YYYYWwwD
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return DateResolution.Weekday
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else:
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raise ISOFormatError('"{0}" is not a valid ISO 8601 week date.'.format(isodatestr))
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#If the size of the string of 4 or less, assume its a truncated year representation
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if len(isodatestr) <= 4:
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return DateResolution.Year
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#An ISO string may be a calendar represntation if:
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# 1) When split on a hyphen, the sizes of the parts are 4, 2, 2 or 4, 2
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# 2) There are no hyphens, and the length is 8
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datestrsplit = isodatestr.split('-')
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#Check case 1
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if len(datestrsplit) == 2:
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if len(datestrsplit[0]) == 4 and len(datestrsplit[1]) == 2:
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return DateResolution.Month
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if len(datestrsplit) == 3:
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if len(datestrsplit[0]) == 4 and len(datestrsplit[1]) == 2 and len(datestrsplit[2]) == 2:
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return DateResolution.Day
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#Check case 2
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if len(isodatestr) == 8 and isodatestr.find('-') == -1:
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return DateResolution.Day
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#An ISO string may be a ordinal date representation if:
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# 1) When split on a hyphen, the sizes of the parts are 4, 3
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# 2) There are no hyphens, and the length is 7
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#Check case 1
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if len(datestrsplit) == 2:
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if len(datestrsplit[0]) == 4 and len(datestrsplit[1]) == 3:
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return DateResolution.Ordinal
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#Check case 2
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if len(isodatestr) == 7 and isodatestr.find('-') == -1:
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return DateResolution.Ordinal
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#None of the date representations match
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raise ISOFormatError('"{0}" is not an ISO 8601 date, perhaps it represents a time or datetime.'.format(isodatestr))
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def parse_date(isodatestr):
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#Given a string in any ISO 8601 date format, return a datetime.date
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#object that corresponds to the given date. Valid string formats are:
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#
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#Y[YYY]
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#YYYY-MM-DD
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#YYYYMMDD
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#YYYY-MM
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#YYYY-Www
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#YYYYWww
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#YYYY-Www-D
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#YYYYWwwD
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#YYYY-DDD
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#YYYYDDD
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#
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#Note that the ISO 8601 date format of ±YYYYY is expressly not supported
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return _RESOLUTION_MAP[get_date_resolution(isodatestr)](isodatestr)
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def _parse_year(yearstr):
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#yearstr is of the format Y[YYY]
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#
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#0000 (1 BC) is not representible as a Python date so a ValueError is
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#raised
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#
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#Truncated dates, like '19', refer to 1900-1999 inclusive, we simply parse
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#to 1900-01-01
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#
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#Since no additional resolution is provided, the month is set to 1, and
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#day is set to 1
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if len(yearstr) == 4:
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isoyear = int(yearstr)
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else:
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#Shift 0s in from the left to form complete year
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isoyear = int(yearstr.ljust(4, '0'))
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if isoyear == 0:
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raise YearOutOfBoundsError('Year must be between 1..9999.')
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return datetime.date(isoyear, 1, 1)
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def _parse_calendar_day(datestr):
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#datestr is of the format YYYY-MM-DD or YYYYMMDD
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if len(datestr) == 10:
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#YYYY-MM-DD
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strformat = '%Y-%m-%d'
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elif len(datestr) == 8:
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#YYYYMMDD
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strformat = '%Y%m%d'
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else:
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raise ISOFormatError('"{0}" is not a valid ISO 8601 calendar day.'.format(datestr))
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parseddatetime = datetime.datetime.strptime(datestr, strformat)
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#Since no 'time' is given, cast to a date
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return parseddatetime.date()
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def _parse_calendar_month(datestr):
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#datestr is of the format YYYY-MM
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if len(datestr) != 7:
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raise ISOFormatError('"{0}" is not a valid ISO 8601 calendar month.'.format(datestr))
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parseddatetime = datetime.datetime.strptime(datestr, '%Y-%m')
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#Since no 'time' is given, cast to a date
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return parseddatetime.date()
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def _parse_week_day(datestr):
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#datestr is of the format YYYY-Www-D, YYYYWwwD
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#
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#W is the week number prefix, ww is the week number, between 1 and 53
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#0 is not a valid week number, which differs from the Python implementation
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#
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#D is the weekday number, between 1 and 7, which differs from the Python
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#implementation which is between 0 and 6
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isoyear = int(datestr[0:4])
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gregorianyearstart = _iso_year_start(isoyear)
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#Week number will be the two characters after the W
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windex = datestr.find('W')
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isoweeknumber = int(datestr[windex + 1:windex + 3])
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if isoweeknumber == 0 or isoweeknumber > 53:
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raise WeekOutOfBoundsError('Week number must be between 1..53.')
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if datestr.find('-') != -1 and len(datestr) == 10:
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#YYYY-Www-D
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isoday = int(datestr[9:10])
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elif len(datestr) == 8:
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#YYYYWwwD
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isoday = int(datestr[7:8])
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else:
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raise ISOFormatError('"{0}" is not a valid ISO 8601 week date.'.format(datestr))
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if isoday == 0 or isoday > 7:
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raise DayOutOfBoundsError('Weekday number must be between 1..7.')
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return gregorianyearstart + datetime.timedelta(weeks=isoweeknumber - 1, days=isoday - 1)
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def _parse_week(datestr):
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#datestr is of the format YYYY-Www, YYYYWww
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#
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#W is the week number prefix, ww is the week number, between 1 and 53
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#0 is not a valid week number, which differs from the Python implementation
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isoyear = int(datestr[0:4])
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gregorianyearstart = _iso_year_start(isoyear)
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#Week number will be the two characters after the W
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windex = datestr.find('W')
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isoweeknumber = int(datestr[windex + 1:windex + 3])
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if isoweeknumber == 0 or isoweeknumber > 53:
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raise WeekOutOfBoundsError('Week number must be between 1..53.')
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return gregorianyearstart + datetime.timedelta(weeks=isoweeknumber - 1, days=0)
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def _parse_ordinal_date(datestr):
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#datestr is of the format YYYY-DDD or YYYYDDD
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#DDD can be from 1 - 36[5,6], this matches Python's definition
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isoyear = int(datestr[0:4])
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if datestr.find('-') != -1:
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#YYYY-DDD
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isoday = int(datestr[(datestr.find('-') + 1):])
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else:
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#YYYYDDD
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isoday = int(datestr[4:])
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parseddate = datetime.date(isoyear, 1, 1) + datetime.timedelta(days=isoday - 1)
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#Enforce ordinal day limitation
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#https://bitbucket.org/nielsenb/aniso8601/issues/14/parsing-ordinal-dates-should-only-allow
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if isoday == 0 or parseddate.year != isoyear:
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raise DayOutOfBoundsError('Day of year must be from 1..365, 1..366 for leap year.')
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return parseddate
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def _iso_year_start(isoyear):
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#Given an ISO year, returns the equivalent of the start of the year
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#on the Gregorian calendar (which is used by Python)
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#Stolen from:
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#http://stackoverflow.com/questions/304256/whats-the-best-way-to-find-the-inverse-of-datetime-isocalendar
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#Determine the location of the 4th of January, the first week of
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#the ISO year is the week containing the 4th of January
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#http://en.wikipedia.org/wiki/ISO_week_date
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fourth_jan = datetime.date(isoyear, 1, 4)
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#Note the conversion from ISO day (1 - 7) and Python day (0 - 6)
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delta = datetime.timedelta(fourth_jan.isoweekday() - 1)
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#Return the start of the year
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return fourth_jan - delta
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_RESOLUTION_MAP = {
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DateResolution.Day: _parse_calendar_day,
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DateResolution.Ordinal: _parse_ordinal_date,
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DateResolution.Month: _parse_calendar_month,
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DateResolution.Week: _parse_week,
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DateResolution.Weekday: _parse_week_day,
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DateResolution.Year: _parse_year
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}
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